Is it possible to create a wrapper around an &mut that acts like an &mut

Question

The following code fails to compile because MutRef is not Copy. It can not be made copy because &\'a mut i32 is not Copy. Is there any way give

Answer 1

No.

The name for this feature is "implicit reborrowing" and it happens when you pass a &mut reference where the compiler expects a &mut reference of a possibly different lifetime. The compiler only implicitly reborrows when the actual type and the expected type are both &mut references. It does not work with generic arguments or structs that contain &mut references. There is no way in current Rust to make a custom type that can be implicitly reborrowed.

However, you can implement your own method to explicitly reborrow:

impl<'a> MutRef<'a> {
    // equivalent to fn reborrow(&mut self) -> MutRef<'_>
    fn reborrow<'b>(&'b mut self) -> MutRef<'b> {
        MutRef {v: self.v}
    }
}

fn wrapper() {
    let mut s: i32 = 9;
    let mut q = MutRef{ v: &mut s };
    wrapper_use(q.reborrow());  // does not move q
    wrapper_use(q);             // moves q
}

See also

• Why is the mutable reference not moved here?
• Type inference and borrowing vs ownership transfer